If 1 x − 1 y = 1 z , d = gcd ( x , y , z ) then d x y z and d ( y − x ) are squaresLet x , y , z be three non negative integer such that 1 x − 1 y = 1 z . Denote by d the greatest common divisor of x , y , zProve that d x y z and d ( y − x ) are squares number.My idea is that if x = d a , y = d b , z = d c, where gcd ( a , b , c ) = 1, then c ( b − a ) = b a. But I don't know how to do next.

Question

If       1    x    −      1    y    =      1    z  ,   d  =  gcd  (  x  ,  y  ,  z  ) then   d  x  y  z and   d  (  y  −  x  ) are squaresLet   x  ,  y  ,  z be three non negative integer such that             1      x        −            1      y        =            1      z      . Denote by   d the greatest common divisor of   x  ,  y  ,  zProve that   d  x  y  z and   d  (  y  −  x  ) are squares number.My idea is that if   x  =  d  a  ,  y  =  d  b  ,  z  =  d  c, where   gcd  (  a  ,  b  ,  c  )  =  1, then   c  (  b  −  a  )  =  b  a. But I don&#039;t know how to do next.

stacan6t · Accepted Answer

It suffices to prove that   d  (  y  −  x  ) is a square, because                    (id)                    (        y        −        x        )        z        =        x        y        ,            and thus  d  (  y  −  x  )      z          2        =  d  x  y  z  .Let   p be a prime, and let       p          a        ,      p          b        ,      p          c       be the highest powers of   p that divide respectively   x  ,  y  ,  zLet us look at (the exponent of) the highest powers of   p that divides   y  −  xIf   a  ≠  b, the highest power of   p that divides   y  −  x is   min  (  a  ,  b  ). Comparing the powers of   p in (id) we get thus  min  (  b  ,  a  )  +  c  =  a  +  b  ,so that   c  =  max  (  a  ,  b  )This implies that the highest power of   p that divides   d  (  y  −  x  ) is  min  (  a  ,  b  ,  c  )  +  min  (  b  ,  a  )  =  2  min  (  b  ,  a  )  .So let us consider the case   a  =  b. Writing   x  =      p          a            x    ′  ,   y  =      p          a            y    ′  , we obtain      p          a        (      y    ′    −      x    ′    )  z  =      x    ′        y    ′        p          2      a        .Therefore, if   e is the highest power of   p that divides       y    ′    −      x    ′  , we have   e  +  c  =  a, so that   c  ≤  a. Thus in this case the highest power of   p that divides   d  (  y  −  x  ) is   c  +  a  +  e  =  2  aWe have proved that the highest power of every prime that divides   d  (  y  −  x  ) has an even exponent. Thus   d  (  y  −  x  ) is a square.

If 1 x </mfrac> − 1 y </mfrac> =

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